Revision as of 11:20, 21 May 2019

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Introduction

The quantum world is governed by the Schrödinger equation

\[{\displaystyle {\hat {H}}|\psi (t)\rangle =i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle } \]

where $\hat H$ is the Hamiltonian, $|\psi (t)\rangle$ is the quantum state function and $\hbar$ is the reduced Planck constant.

The Hamiltonian consists of kinetic energy $\hat T$ and potential energy $\hat V$. As in classical mechanics, potential energy is a function of time and space, whereas the kinetic energy differs from the classical world and is calculated as

\[\hat T = - \frac{\hbar^2}{2m} \nabla^2 .\]

The final version of the single particle Schrödinger equation can be written as

\[\left(- \frac{\hbar^2}{2m} \nabla^2 + V(t, \mathbf r)\right) \psi(t, \mathbf r) = i\hbar {\frac {\partial }{\partial t}}\psi(t, \mathbf r) \]

Quantum state function is a complex function, so it is usually split into the real part and imaginary part

\[ u, v \in C(\mathbb R)\colon \psi = u + i v , \]

which for a real $V$ yields a system of two real equations

\[\left(- \frac{\hbar^2}{2m} \nabla^2 + V(t, \mathbf r)\right) u(t, \mathbf r) = -\hbar {\frac {\partial }{\partial t}} v(t, \mathbf r) , \] \[\left(- \frac{\hbar^2}{2m} \nabla^2 + V(t, \mathbf r)\right) v(t, \mathbf r) = \hbar {\frac {\partial }{\partial t}} u(t, \mathbf r) , \]

which may be easier to handle.

Particle in a box

By selecting the potential $V(t, \mathbf r)$ and the initial state $\psi(0, \mathbf r)$ we get a unique solution for time propagation of the quantum state function. A theoretical one dimensional potential

\[\displaystyle V(x)={\begin{cases}0,&0<x<L,\\\infty ,&{\text{otherwise,}}\end{cases}}\]

is known as an infinite potential well. Its time independent eigenfunctions are

\[\sqrt{\frac{2}{L}}\psi_n(x) = \sin\left(k_n x \right), \qquad n = 1,2,3,...\]

where $k_n = \frac{\pi n}{L}$. With a time dependency

\[\psi_n(t, x) = \mathrm e ^ {-i \omega_n t} \psi_n(x),\]

where $\omega_n$ and $k_n$ are connected through dispersion relation through energy $E_n$

\[{\displaystyle E_{n}=\hbar \omega _{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}={\frac {\hbar ^{2} k_n^2}{2m}}}.\]